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Optimization Problem - Calculus

  • Thread starter polak333
  • Start date
  • #1
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Homework Statement



A cylindrical shaped tin can must have a volume of 1000cm3.

Find the dimensions that require the minimum amount of tin for the can (Assume no waste material). The smallest can has a diameter of 6cm and a height of 4 cm.

Homework Equations



[tex]V = \pi r^{2}h[/tex]

[tex]P = 2( \pi r^{2}) + (2 \pi r)h[/tex]

The Attempt at a Solution



I basically start off by isolating a variable:

[tex]V = \pi r^{2}h[/tex]

[tex]1000 = \pi r^{2}h[/tex]

[tex]318.3 = r^{2}h[/tex]

[tex]r^{2} = \frac{318.3}{h}[/tex]
------------------------------------------------------------------
This is where I'm stuck. Do I plug this into the surface area formula.
I know I have to sub this into something else, and then expand and find the derivative of that new equation to find where h > 4.
Just need to find out what to do next from here.

Thanks for any help!
 

Answers and Replies

  • #2
33,515
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You have a relationship between r and h (but leave pi in - 1000/pi is not exactly equal to 318.8). Now rewrite your surface area function P (why is it P?) so that it is a function of one variable, either r or h. Then use the normal technique for finding the minimum or maximum function value.
 

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